A circular disc is rolling on a horizontal pl | vedclass

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Question

Total $\mathrm{KE}=\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right) \frac{1}{2} \mathrm{mv}^{2}$

Translational $\mathrm{KE}=\frac{1}{2} \mathrm

Vedclass · Accepted Answer

$200$

Vedclass · Answer

Total $\mathrm{KE}=\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}ight) \frac{1}{2} \mathrm{mv}^{2}$

Translational $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}$

$\frac{	ext { Trans } \mathrm{KE}}{	ext { Total } \mathrm{KE}}=\frac{1}{1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}}$

Here, $\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{1}{2} \Rightarrow$ Trans $\mathrm{KE}=\frac{1}{1+\frac{1}{2}}$ Total $\mathrm{KE}$

Trans $\mathrm{KE}=\frac{2}{3} 	imes 300=200 \mathrm{J}$

A circular disc is rolling on a horizontal plane. Its total kinetic energy is $300\,J$ . ....... $J$ is its translational $K.E$

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